Import Question JSON

Current Question (ID: 18042)

Question:

$\text{A solution of glucose (M.W = 180 g mol}^{-1}\text{) in water is 10 \% w/w. The mole fraction of each component in the solution is:}$

Options:

1. $\text{Mole fraction of glucose - 0.44; mole fraction of water - 0.56}$
2. $\text{Mole fraction of glucose - 0.056; mole fraction of water - 0.944}$
3. $\text{Mole fraction of glucose - 0.011; mole fraction of water - 0.99}$
4. $\text{Mole fraction of glucose - 0.36; mole fraction of water - 0.64}$

Solution:

$\text{Hint: } X_{\text{mole fraction}} = \frac{\text{number of mole of solute}}{\text{total mole of solution}}$ $\text{Step 1:}$ $\text{Find the number of moles of solute (glucose) and solvent (water).}$ $10\% \text{ w/w means 10 grams of glucose present in 100 g of solution. The amount of solvent is 90 g}$ $\text{Number of mole of solute (glucose)} = \frac{\text{amount of glucose}}{\text{molar mass of glucose}}$ $\text{Number of mole of solute (glucose)} = \frac{10 \text{ g}}{180 \text{ g mol}^{-1}}$ $\text{Number of mole of solute (glucose)} = 0.056 \text{ mol}$ $\text{Number of mole of solvent (water)} = \frac{\text{amount of water}}{\text{molar mass of water}}$ $\text{Number of mole of solvent (water)} = \frac{90 \text{ g}}{18 \text{ g mol}^{-1}}$ $\text{Number of mole of solvent (water)} = 5 \text{ mol}$ $\text{Step 2:}$ $\text{Find the mole fraction of each component}$ $X_{\text{glucose}} = \frac{\text{number of mole of glucose}}{\text{total mole of solution}}$ $X_{\text{glucose}} = \frac{0.056}{0.056 + 5}$ $X_{\text{glucose}} = 0.011$ $X_{\text{water}} = 1 - 0.011$ $X_{\text{water}} = 0.989 \approx 0.99$ $X_{\text{water}} = 0.99$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}